Question Number 208111 by mr W last updated on 05/Jun/24
Commented by mr W last updated on 05/Jun/24
$${yellow}\:{area}\:=? \\ $$
Answered by A5T last updated on 05/Jun/24
Commented by A5T last updated on 05/Jun/24
$$\frac{\mathrm{1}}{\mathrm{2}}\left({s}+{AB}\right){t}=\mathrm{14}\Rightarrow{AB}=\frac{\mathrm{28}}{{t}}−{s} \\ $$$$\left({t}−\frac{\mathrm{16}}{{s}}\right)×\frac{{BC}}{\mathrm{2}}=\mathrm{2}\Rightarrow{BC}=\frac{\mathrm{4}{s}}{{st}−\mathrm{16}} \\ $$$$\left({AB}+{BC}\right)×\frac{{t}}{\mathrm{2}}=\mathrm{8}\Rightarrow{AB}+{BC}=\frac{\mathrm{16}}{{t}} \\ $$$$\Rightarrow\left(\frac{\mathrm{28}−{st}}{{t}}+\frac{\mathrm{4}{s}}{{st}−\mathrm{16}}\right)=\frac{\mathrm{16}}{{t}} \\ $$$$\Rightarrow\mathrm{28}−{st}+\frac{\mathrm{4}{st}}{{st}−\mathrm{16}}=\mathrm{16} \\ $$$$\Rightarrow\left({st}\right)^{\mathrm{2}} −\mathrm{32}{st}+\mathrm{192}=\mathrm{0}\Rightarrow{st}=\mathrm{8}\:{or}\:{st}=\mathrm{24} \\ $$$$\Rightarrow{Yellow}\:{area}=\mathrm{24}−\mathrm{8}−\mathrm{6}−\mathrm{2}=\mathrm{8} \\ $$
Commented by mr W last updated on 05/Jun/24
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Answered by mr W last updated on 05/Jun/24
Commented by mr W last updated on 05/Jun/24
$$\frac{\mathrm{2}}{\mathrm{8}}=\left(\frac{{y}}{{a}}\right)^{\mathrm{2}} \:\Rightarrow\frac{{y}}{{a}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{6}+\mathrm{2}}{\mathrm{6}+\mathrm{8}}=\frac{{x}+{y}}{{x}+{a}}\:\Rightarrow\frac{{x}+{y}}{{x}+{a}}=\frac{\mathrm{4}}{\mathrm{7}}\:\Rightarrow\frac{{x}}{{a}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\frac{{z}}{{a}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\frac{{A}+\mathrm{2}}{\mathrm{6}+\mathrm{2}}=\frac{{z}+{y}}{{x}+{y}}=\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{5}}{\mathrm{4}}×\mathrm{8}−\mathrm{2}=\mathrm{8}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/24
$$\mathrm{weldone}\:\mathrm{sir} \\ $$
Answered by A5T last updated on 05/Jun/24
Commented by A5T last updated on 05/Jun/24
$$\left({t}−\frac{\mathrm{16}}{{s}}\right)×\frac{{BC}}{\mathrm{2}}=\mathrm{2}\Rightarrow{BC}=\frac{\mathrm{4}{s}}{{st}−\mathrm{16}} \\ $$$$\frac{\mathrm{2}}{\mathrm{8}}=\frac{{BC}^{\mathrm{2}} }{{s}^{\mathrm{2}} }\Rightarrow{BC}=\frac{{s}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{4}{s}}{{st}−\mathrm{16}}=\frac{{s}}{\mathrm{2}}\Rightarrow{st}=\mathrm{24} \\ $$$$\Rightarrow{Yellow}\:{area}=\mathrm{24}−\mathrm{8}−\mathrm{6}−\mathrm{2}=\mathrm{8} \\ $$